Two ideal gases A and B of the same number of | vedclass

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(D) The work done in an isothermal process $(T = 	ext{constant})$ is given by $W = nRT \ln \left( \frac{P_i}{P_f} ight)$.For gas $A$, the pressure

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$2: 1$

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(D) The work done in an isothermal process $(T = 	ext{constant})$ is given by $W = nRT \ln \left( \frac{P_i}{P_f} ight)$.For gas $A$, the pressure decreases by $50 \%$, so $P_f = P_i - 0.5 P_i = 0.5 P_i$. Thus, $\frac{P_i}{P_f} = \frac{1}{0.5} = 2$.For gas $B$, the pressure decreases by $75 \%$, so $P_f = P_i - 0.75 P_i = 0.25 P_i$. Thus, $\frac{P_i}{P_f} = \frac{1}{0.25} = 4$.Given that the work done by both gases is equal $(W_1 = W_2)$ and the number of moles $(n)$ is the same:$nRT_1 \ln(2) = nRT_2 \ln(4)$$T_1 \ln(2) = T_2 \ln(2^2)$$T_1 \ln(2) = 2 T_2 \ln(2)$Dividing both sides by $\ln(2)$:$T_1 = 2 T_2 \Rightarrow \frac{T_1}{T_2} = \frac{2}{1}$.Therefore, the ratio $T_1: T_2$ is $2: 1$.

Two ideal gases $A$ and $B$ of the same number of moles expand at constant temperatures $T_1$ and $T_2$ respectively such that the pressure of gas $A$ decreases by $50 \%$ and the pressure of gas $B$ decreases by $75 \%$. If the work done by both the gases is same, then $T_1: T_2$ is:

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